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Shortcuts for Projections Problem: Show that these formulas give the scalar and vector projections of \(\vec{a}\) on \(\vec{b}\) The scalar projection of \(\vec{a}\) on \(\vec{b}\) is given by \(p=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) The vector projection of \(\vec{a}\) on \(\vec{b}\) is given by \(\vec{p}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^{2}} \vec{b}\)

Short Answer

Expert verified
The scalar and vector projections formulas are derived using the vector dot product and vector magnitude, resulting in \(p\) as the scalar projection and \(\vec{p}\) as the vector projection of \(\vec{a}\) on \(\vec{b}\).

Step by step solution

01

Set Up The Problem

Understand that the problem involves finding the scalar and vector projections of one vector onto another. The scalar projection is a measure of how much one vector extends in the direction of another vector, and the vector projection is the vector that represents that extension.
02

Review Vector Dot Product

Recall that the dot product of two vectors \(\vec{a} \cdot \vec{b}\) gives a scalar value and it is calculated by multiplying corresponding components of the two vectors and then summing those products.
03

Review Vector Magnitude

Recall that the magnitude of a vector \(\vec{b}\) is denoted by \( |\vec{b}| \) and is calculated by taking the square root of the sum of the squares of its components.
04

Compute Scalar Projection

Using the formula \( p = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \), calculate the scalar projection by dividing the dot product of \(\vec{a}\) and \(\vec{b}\) by the magnitude of \(\vec{b}\).
05

Compute Vector Projection

Using the formula \( \vec{p} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^{2}} \vec{b} \), calculate the vector projection by multiplying the scalar projection by the unit vector in the direction of \(\vec{b}\), which is \(\vec{b} / |\vec{b}|\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Projection Formula
The scalar projection formula is a critical component when studying vectors in a geometric context. In simple terms, it measures how far a given vector, let's say \( \vec{a} \), extends along the direction of another vector \( \vec{b} \). It's essentially the 'shadow' or 'footprint' of one vector onto another when the light source is perpendicular to the second vector.

To calculate the scalar projection, represented by \( p \), you use the formula \( p = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \). Here, the dot product \( \vec{a} \cdot \vec{b} \) quantifies how much the direction of \( \vec{a} \) aligns with \( \vec{b} \) by considering their lengths and the angle between them. This result is then divided by the magnitude of \( \vec{b} \) to adjust for its length, leaving you with the length of the 'shadow' we mentioned.

Remember to view the scalar projection as a way to gauge the influence of one vector along the direction of another, which has practical usage in physics for understanding forces or in computer graphics for calculating light intensities.
Vector Dot Product
The vector dot product, also known as the inner product or scalar product, is a fundamental operation in vector algebra. It's used to determine the magnitude of one vector in the direction of another. To find the dot product of two vectors \( \vec{a} \) and \( \vec{b} \), you multiply their corresponding components and then add these products together, mathematically expressed as \( \vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z \) for three-dimensional vectors.

One crucial property of the dot product is that it can tell you about the angle between the two vectors. If the dot product is positive, the vectors are oriented in the same general direction. If it's zero, the vectors are perpendicular, and if it's negative, they are pointing in generally opposite directions.

This operation is not just a mathematical convenience but a powerful tool in physics for work and energy calculations as well as in computer science for rendering 3D models, where it helps in determining how light reflects off surfaces.
Vector Magnitude
Understanding vector magnitude is akin to gauging the 'size' or 'length' of a vector. The magnitude of a vector \( \vec{b} \) is denoted by \( |\vec{b}| \), and it's calculated by finding the square root of the sum of the squares of its components. For a two-dimensional vector with components \( b_x \) and \( b_y \) this is expressed as \( |\vec{b}| = \sqrt{b_x^2 + b_y^2} \).

In the physical world, the magnitude of a vector can represent speed, force, or other quantities depending on the context. It's measured in units that depend on what the vector itself represents. For instance, if a vector describes a force, its magnitude will be in Newtons.

Always remember that the magnitude provides essential information about the vector's 'strength' without considering its direction, which differentiates it from variables like velocity that provide both magnitude and direction information.
Vector Projection Formula
While a scalar projection gives you the length of a vector's shadow on another vector, the vector projection formula gives you the actual vector lying along the second vector that represents this projection. The formula is given by \( \vec{p} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \).

To compute this, you first perform the dot product of \( \vec{a} \) and \( \vec{b} \), and divide it by the square of the magnitude of \( \vec{b} \). This scalar quantity is then multiplied with the vector \( \vec{b} \) itself. The vector \( \vec{b} \) is usually normalized first, meaning it's been reduced to a unit vector in the same direction as \( \vec{b} \) but with a magnitude of 1. This projection is akin to casting the entirety of \( \vec{a} \) onto the line of \( \vec{b} \) so you can see its full influence in that specific direction.

The vector projection is used in many fields, including physics where it can, for example, show the specific direction a force is applied in relation to another direction, or in engineering for resolving components of vectors along particular axes.

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Most popular questions from this chapter

Flood Control Tunnel Problem: Suppose that you work for a construction company that has been hired to dig a drainage tunnel under a city. The tunnel will carry excess water to the other side of the city during heavy rains, thus preventing flooding (Figure \(10-8 f\) ). The tunnel is to start at ground level and then slant down until it reaches a point \(100 \mathrm{ft}\) below the surface. Then it will go horizontally, far enough to reach the other side of the city (not shown). Your job is to analyre the slanted part of the tunnel. GRAPH CANT COPY a. The Engineering Department has determined that the tunnel will slant downward in the direction of the vector \(\vec{v}=97+127-20 \vec{k}\) The centerline of the tunnel starts at the point (30,40,0) on the surface. The measurements are in feet. Write the particular vector equation of the centerline. b. How far along the centerline must the construction crews dig to reach the end of the slanted part of the tunnel, \(100 \mathrm{ft}\) below ground? What are the coordinates of this endpoint? c. Construction crews must be careful when they reach a fault plane that is in the path of the slanted part of the tunnel. The Geology Department has determined that the point (60,90,0) is on the fault plane where it outcrops at ground level and that the vector \(\vec{n}=2 \vec{i}-4 \vec{j}+\vec{k}\) is normal to the plane (Figure \(10-8 f\) ). Find the particular equation of the plane. d. How far along the centerline of the tunnel must the construction crews dig in order to reach the fault plane? What are the coordinates of the point at which the centerline intersects the plane? How far beneath ground level is this point?

Prove that the vector is a unit vector, and find its direction cosines and direction angles. $$\frac{1}{9} \vec{\imath}-\frac{4}{9} \vec{\jmath}+\frac{8}{9} \vec{k}$$

Find a particular equation of the plane described. Perpendicular to \(\vec{n}=3 \vec{\imath}-5 \vec{\jmath}+4 \vec{k},\) containing the point (6,-7,-2)

Find a particular equation of the plane containing the given points. $$(0,3,-7),(5,0,-1), \text { and }(4,3,9)$$

Find the indicated displacement vector. Use the answer to find the distance between the two points. \(\overrightarrow{B A}\) for \(A(9,13,-4)\) and \(B(3,6,-10)\)

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